Question:

Step 1: To find the number of triangles that can be formed using the points $P_1,P_2,\ldots,P_{18}$ as vertices, we first need to calculate the total number of combinations of 3 points from the 18 points. This can be calculated using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items and $r$ is the number of items to choose. In this case, we have $n=18$ and $r=3$, so the total number of combinations is $\binom{18}{3} = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$. Step 2: However, not all combinations of 3 points will form valid triangles. Specifically, if all 3 points are on the same side of the triangle, they will be collinear and cannot form a triangle. We need to subtract the number of such combinations from the total. There are 5 points on side $AB$, so the number of combinations of 3 points on this side is $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$. Similarly, there are 6 points on side $BC$, so the number of combinations of 3 points on this side is $\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. Finally, there are 7 points on side $CA$, so the number of combinations of 3 points on this side is $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. Step 3: To find the total number of valid triangles, we subtract the number of combinations that do not form triangles from the total number of combinations. This gives us $\binom{18}{3} - \binom{5}{3} - \binom{6}{3} - \binom{7}{3} = 816 - 10 - 20 - 35 = 751$. Step 4: Therefore, the number of triangles that can be formed using the points $P_1,P_2,\ldots,P_{18}$ as vertices is $751$, which corresponds to Option 3. The final answer is $\boxed{751}$, and this matches Option 3.